Separable Differential Equations

What is separable differential equation?

A differential Equation is separable if it can be written as f(y) dy= g(x) dx. Its solution can be found by integration both sides. A differential equation where we are able to move all y terms and dy to left and all x terms along with dx to right side, is called separable differential equation.

For example

$\frac{dy}{dx}= x siny$ is separable but

$\frac{dy}{dx}=x+y$ is not separable because x+y is not a product f(y) g(x) so can’t be separated.

Example1: Solve the given differential equation.

${\color{Red} \frac{dy}{dx}=\frac{2xy}{x+1}}$

Solution: Here we can easily separate the variables because right side is a product of x and y terms where f(y) =y and g(x)=2x/(x+1)

Separating the variables, we get

$\frac{1}{y} dy=\frac{2x}{x+1}dx$

Integrating both sides,

$\int \frac{1}{y} dy=\int \frac{2x}{x+1}dx$

$\int \frac{1}{y} dy=2\int \frac{x+1-1}{x+1}dx$

$\int \frac{1}{y} dy=2\int 1-\frac{1}{x+1}dx$

ln|y|=2[x-ln|x+1|]+C

Example2:Solve the differential equation
${\color{Red} \frac{dy}{dx}=\sqrt{16x^{2}y-4x^2y^2}}$ , y(2)=1

Solution: After simplifying the right side to separate x and y terms, we get

$\frac{dy}{dx}=\sqrt{4x^{2}(4y-y^2)}$

$\frac{dy}{dx}=2x\sqrt{4y-y^2}$

$\frac{1}{\sqrt{4y-y^2}}dy=2x dx$

$\int \frac{1}{\sqrt{4-4+4y-y^2}}dy=\int 2x dx$

$\int \frac{1}{\sqrt{2^2-(y-2)^2}}dy=\int 2x dx$

$sin^{-1}\left ( \frac{y-2}{2} \right )=x^{2}+C$

Using given initial condition,plugin x=2 and y=1 to get value of C

$sin^{-1}\left ( \frac{1-2}{2} \right )=2^{2}+C$

$sin^{-1}\left ( \frac{-1}{2} \right )-4=C$

$\frac{7\pi }{6}-4=C$

So the final solution is given as,

$sin^{-1}\left ( \frac{y-2}{2} \right )=x^{2}+ \frac{7\pi }{6}-4$

Example3: A parachutist falling toward Earth is subject to two forces: the parachutist weight (w = 32m) and the drag of the parachute. The drag of the parachute is proportional to the velocity of the parachute and in this case is equal to 8|v|. The parachutist weight is 128lb and initial velocity is zero. Find formulas for the parachutist’s velocity v(t) and distance x(t). Also find parachutists terminal velocity.

Solution: Since the parachutist falls down (the positive direction) velocity is always positive so |v| = v. The resultant force will be the force of the weight of the parachutist minus the force of the drag of the parachute.

so Force F= 128-8v ……. (i)

And the mass of parachute is, 128= 32m

therefore, m=4

Using Newton’s second law, Force F=ma

since a= dv/dt ,

so F=4 dv/dt …….. (ii)

Using equations (i) and (ii) we get,

$4\frac{dv}{dt}=128-8v$

$\int \frac{1}{32-2v}dv =\int dt$

$\frac{1}{2}\int \frac{1}{16-v}dv =\int dt$

$\frac{-1}{2}ln|16-v| = t+C$

$ln|16-v| = -2t-2C$

$ln|16-v| = -2t-K$

Using initial condition v(0)=0 , we get

-ln(16)=K

$ln|16-v| = -2t+ln(16)$

ln|16-v|-ln(16) =-2t

$ln\left ( \frac{16-v}{16} \right )=-2t$

$\left ( \frac{16-v}{16} \right )=e^{-2t}$

Solving for v , we get

$v(t)=16-16e^{-2t}$

Terminal velocity

$\lim_{t\rightarrow \infty }v(t)=\lim_{t\rightarrow \infty }(16-16e^{-2t})=16-0=16$ ft/sec

To know how far parachute has fallen, we find position function x(t) by taking integral of velocity function.

x(t)=∫v(t) dt

$x(t)=\int (16-16e^{-2t}) dt$

$x(t)=16t+8e^{-2t}+C$

using initial condition x(0)=0 we get

x(0)=0+8+C

-8=C

$x(t)=16t+8e^{-2t}-8$

Example4. Suppose that a cup of coffee is initially at a temperature of 105° F and is placed in a 75° F room. Newton’s law of cooling says that

${\color{Red} \frac{dT}{dt}=-k(T-75)}$

where $k$ is a constant of proportionality.

a) Suppose you measure that the coffee is cooling at one degree per minute at the time the coffee is brought into the room. Use the differential equation to determine the value of the constant $k$

b)Find all the solutions of this differential equation.

c)What happens to all the solutions as $t \to \infty$ ? Explain how this agrees with your intuition.

d) What is the temperature of the cup of coffee after 20 minutes?

e) How long does it take for the coffee to cool to 80°?

Solution:

a) Given dT/dt= -1 when T=105

Using these values into the differential equation, we get

-1=-k(105-75)

1/30 =k

b) Using this value of k we get the original DE as,

$\frac{dT}{dt}=-\frac{1}{30}(T-75)$

$\frac{dT}{T-75}=\frac{-1}{30}dt$

$\int \frac{dT}{T-75}=\frac{-1}{30}\int dt$

$ln|T-75|=-\frac{1}{30}t+c$

$T-75=e^{\frac{-1}{30}t+c}$

$T=75+e^{\frac{-1}{30}t}*A$

$T=75+Ae^{\frac{-t}{30}}$

Using initial condition T(0)=105, we get

$T(0)=75+Ae^{0}$

105-75=A => A=30

$T(t)=75+30e^{\frac{-t}{30}}$

$\lim_{t\rightarrow \infty }T(t)= \lim_{t\rightarrow \infty }[75+Ae^\frac{-t}{30}]=75+A(0)=75$

When t approaches to infinity , exponential part become 0 and the temp of coffee is same as temp of room.

d) After 20 minutes

$T(20)=75+30e^{\frac{-20}{30}}$

T(20)=90.4ºF

e) given T(t)=80

$80=75+30e^{\frac{-t}{30}}$

$\frac{5}{30}= e^{\frac{-t}{30}}$

ln(1/6)=-t/30

t=-30ln(1/6)

t=53.75 min

Equations reducible to Separable form:

Differential equations of the form dy/dx = f(ax+by+c) can be reduced to variable separable form by substituting ax+by+c= v as discussed in the following example.

Example5:Solve the following differential equation.

${\color{Red} \frac{dy}{dx}=(4x+y+1)^{2}}$

Solution: here we are not able to separate variables x and y so we use substitution.

Let v=4x+y+1

$\frac{dv}{dx}=4+\frac{dy}{dx}$

$\frac{dv}{dx}-4=\frac{dy}{dx}$

Using these values we get the original DE changed to

$\frac{dv}{dx}-4= v^{2}$

$\frac{dv}{dx}= v^{2}+4$

$\frac{1}{v^{2}+4}dv = dx$

$\int \frac{1}{v^{2}+4}dv = \int dx$

$\frac{1}{2}tan^{-1}\left ( \frac{v}{2} \right )=x+C$

Plugin back v to get final answer in terms of x and y

$\frac{1}{2}tan^{-1}\left ( \frac{4x+y+1}{2} \right )=x+C$

Practice problems:

Solve the following differential equations.

1. $\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$

Solve the following initial value problem.

2. $(x+1)\frac{dy}{dx}=2e^{-y}-1 : y(0)=0$

3. $\frac{dy}{dx}=\frac{(x-y)+3}{2(x-y)+5}$

4. A freshly brewed cup of coffee has temperature 95◦C in a 20◦C room. When its temperature is 70◦C, it is cooling at a rate of 1◦C per minute. When does this occur?

Answers:

y-x=C(1+xy)
$y=ln\left ( 2-\frac{1}{x+1} \right )$
2(x-y)+ln(x-y+2)=x+C
20.3 min

What is separable differential equation?

Equations reducible to Separable form:

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